TITLE: A conjecture concerning primes and algebra
QUESTION [16 upvotes]: A monoid morphism $\psi:\mathbb Z_+\!\!\rightarrow\mathbb Z_+$ is defined by an arbitrary function $f:\mathbb Z_+\!\!\rightarrow\mathbb Z_+$ and defines a group homomorphism $\varphi:\mathbb Q_+\!\!\rightarrow\mathbb Q_+$:
$\psi(a\cdot b)=\psi(a)\cdot\psi(b)\Rightarrow\psi(\prod p_i^{n_i})=\prod\psi(p_i)^{n_i}$, so any function $i\mapsto \psi(p_i)$ defines $\psi$. Further, since 
$\varphi(\frac{1}{n})=\frac{1}{\varphi(n)}$ a function $f$ uniquely defines the group homomorphism $\varphi$. 
For fun I intend to study the group homomorphism generated by the identity function 
$i\mapsto\omega(p_i)=i$. Then $\omega:\mathbb Z_+\!\!\rightarrow\mathbb Z_+$ seems to be an algebraic analogue to the analytic approach with $\pi(N)$ since also $\omega$ is a left inverse to the prime number function $p_n$. 
In the diagram $\omega$ is the blue curve and $\pi$ the red and one can see (especially if the picture is loaded and viewed in a larger scale) that $\omega(1)>\pi(1),\; \omega(35)>\pi(35)$ and $\omega(49)>\pi(49)$. Of course $\omega$ and $\pi$ coincide for all primes, but also for other values as $91,\;95$ and $133$. However, I have tested all values less than $2^{16}$ and $\omega(N)\leq\pi(N)$ for $N>49$ in this interval.

I have no idea how to prove the conjecture (if possible) but I will try to construct a program that (in principle) can test any interval of integers.
Edit: I realize now that it is virtually impossible to make such a program, but with help of arrays of primes my program have tested all values $< 1\,000\,000$ and the only numbers for which $\omega(N)\geq\pi(N)$ (italic numbers stands for '$>$') are:
$\mathit{1},\,9,\,15,\,21,\,25,\,\mathit{35},\,39,\,\mathit{49},\,57,\,65,\,91,\,95,\,133$.

Conjecture: For $N=\prod p_k^{n_k}>49$ it holds that $\prod k^{n_k}\leq\pi(N)$.

Thoughtful: from an analytic point of view $\omega$ seems to be very irregular, but algebraically it is an almost canonical morphism.

REPLY [4 votes]: It took a long time before I understood how Greg Martin's lemma
$$N\ge17\wedge n\ge5\implies \pi(p_nN)>n\pi(N)$$
could be exploited. The method in the proof of the lemma required $n\ge5$, but the lemma is true even for $1\le n\le 4$, which can be proved using an inequality of Pierre Dusart proved in 2010:
$\displaystyle \frac {x} {\ln x - 1} < \pi(x)$, for $x \ge 5393$, and
$\displaystyle \pi(x) <  \frac {x} {\ln x - 1.1}$, for $x \ge 60184$. 
E.g. for $n=4$. If $N$ is sufficiently large: $3\ln N>11.7$ why 
$\displaystyle\frac{\ln N+1}{7N}<\frac{\ln N-1.1}{4N}$ which implies that
$\displaystyle\frac{7N}{\ln N+1}>\frac{4N}{\ln N-1.1}.\;$ Now, due to Dusart, $\displaystyle\pi(7N)>\frac{7N}{\ln 7N-1}>\frac{7N}{\ln N+1}>\frac{4N}{\ln N-1.1}>4\pi(N)$. 
All four cases ($n=1,2,3,4$) has been tested on my computer and holds for $N>7$ and $p_nN<1\,000\,000$, which is big enought. $\square$
The general lemma is: $N\geq 17\implies\pi(p_nN)>n\pi(N)$. It should be pointed out that in the proof $N$ has to be larger, but the proof is complemented with computer tests.
Hence, suppose $\omega(x)\leq\pi(x)$ for all $x<M=p_nN$, then if $N\ge17$
$\omega(p_nN)=n\omega(N)\leq n\pi(N)<\pi(p_nN)$. If $N<17$ and $p_n\geq17$ it holds that
$\pi(p_nN)>\pi(p_n)\omega(N)=\omega(p_n)\omega(N)=\omega(p_nN)$. The case with $N<17$ and $p_n<17$ is covered by the computer test $(x>49)$. So the conjecture is proved by induction.